Hilbert class field

The Hilbert class field is the maximal unramified abelian extension of a number field. Its Galois group is canonically the ideal class group.

For a CM-field $K$ over its maximal real subfield $K^{+}$, the ordinary Hilbert class field $H_{K^{+}}$ is totally real because its real places split. Thus $H_{K^{+}}\cap K=K^{+}$. Its compositum with $K$ is an abelian extension of $K$ of degree $h_{K^{+}}$, and is everywhere unramified by unramified extensions remain unramified under base change. It lies in $H_K$, giving the divisibility. The argument does not require injectivity of extension of ideal classes.

The three reduced primitive binary quadratic forms of discriminant $-23$ are $(1,1,6)$ and $(2,\pm 1,3)$, so the class number is three. The displayed cubic has discriminant $-23$ and is irreducible. Its splitting field has Galois group $S_3$ and is cyclic cubic over $\mathbb{Q}(\sqrt{-23})$. Modulo $23$ it factors as $(X-10{)}^2(X-3)$. Lift the simple factor; the remaining quadratic has discriminant valuation one, so its local inertia group has order two. Its intersection with $A_3$ is trivial. Thus the relative cyclic cubic extension is everywhere unramified and equals the Hilbert class field.

The polynomial is irreducible with discriminant $229$, so its splitting field has Galois group $S_3$ and quadratic subfield $\mathbb{Q}(\sqrt{229})$. The splitting field is cyclic cubic over that subfield. It is unramified outside 229, and at 229 its inertia is a transposition, meeting the alternating cubic subgroup trivially. All roots are real. It is therefore everywhere unramified over the quadratic field. The ideal class group of Q of square root 229 has order three, so this extension is exactly the Hilbert class field.

The ideal class group of $\mathbb{Q}(\sqrt{-30})$ is $C_2\times C_2$. Its four reduced binary quadratic forms have coefficients $(1,0,30)$, $(2,0,15)$, $(3,0,10)$ and $(5,0,6)$. The displayed multiquadratic field has degree four over the base. The seven quadratic-subfield discriminants are $8,-3,5,-24,40,-15,-120$, whose product in absolute value is $120^4$. The conductor-discriminant formula and discriminant tower formula imply that the relative discriminant over $\mathbb{Q}(\sqrt{-30})$ is the unit ideal. This unramified abelian degree-four extension is therefore its Hilbert class field.

The Hilbert class field is the maximal everywhere-unramified abelian extension of a number field; its Galois group is canonically the ideal class group.

If a finite extension $B/A$ of number fields has a totally ramified finite prime, the ordinary Hilbert class field $H_A$ is disjoint from $B$. Its base change $B{H}_A/B$ is abelian and unramified at finite primes. Its real places also split, since a split-real unramified extension retains this property under base change. It therefore lies in the ordinary Hilbert class field of $B$, proving $h_A\mid h_B$ even when $A$ has real places.

The splitting field of $T^3+T-1$ has discriminant square class $-31$ and Galois group $S_3$. It is the Hilbert class field of $K=\mathbb{Q}(\sqrt{-31})$: the relative extension is unramified cyclic cubic, and the ideal class group of Q of square root minus thirty-one has order three. Locally at $31$, the simple residue root $28$ lifts and the remaining quadratic splits over ${\mathbb{Q}}_{31}(\sqrt{-31})$, so the prime of $K$ above $31$ splits completely in its Hilbert class field. The inverse generator satisfies $T^3-T^2-1=0$.

Suppose an irreducible monic integral cubic has discriminant $-p$ for a prime $p>3$, and let $L$ be its splitting field. Its Galois group is $S_3$ and its quadratic subfield is $\mathbb{Q}(\sqrt{-p})$. The extension over that subfield is cyclic cubic and everywhere unramified. Away from $p$, separable reduction and Hensel lemma put the roots in an unramified local extension. At $p$, discriminant valuation one gives a double residue root and a separate simple root; lift the simple root, and the remaining quadratic's discriminant has square class $-p$. Its splitting field is already the quadratic subfield's completion. The imaginary quadratic base has no real places.

If $M/L$ is an extension of number fields, $L$ has no real places, and one finite prime of $L$ is totally ramified in $M$, then $h_L\mid h_M$. The Hilbert class field $H_L$ has trivial intersection with $M$: any common intermediate extension would be both unramified and totally ramified at that prime. Thus $M{H}_L/M$ is an unramified abelian extension of degree $h_L$, and is contained in $H_M$.

The splitting field has Galois group $S_3$ and quadratic subfield $\mathbb{Q}(\sqrt{-59})$. The extension over that subfield is abelian of degree three and unramified everywhere. The only possible rational ramified prime is $59$; at that prime a simple residue root lifts, so the local splitting field is quadratic and coincides with the quadratic subfield's completion. It follows from the Hilbert class field degree that $3\mid h_{\mathbb{Q}(\sqrt{-59})}$.

The ideal class group of $K=\mathbb{Q}(\sqrt{-6})$ has order two, andThis quadratic extension of $K$ is unramified at every finite prime and therefore has the degree required of the Hilbert class field.